我把数据放在同一行有问题。
这是我的ui显示：
数据库：
我想创建一个位置，如果用户同时分配腿部和手臂练习，数据库将显示插入一行的腿部和手臂练习。但是现在，当我把它们插入到一起时，数据库中的数据将在两个不同的行中。
.js代码
<%@page import="java.sql.*"%>
<html>
<body>
<%
String[] id = request.getParameterValues("assignchkbox");
String armID = request.getParameter("ArmID");
String armnumber = request.getParameter("Armnumber");
int count=0;
Connection conn = null;
try {
Class.forName("com.mysql.jdbc.Driver");
// Step 2: Define Connection URL
String connURL = "jdbc:mysql://localhost/medicloud?user=root&password=root";
// Step 3: Establish connection to URL
conn = DriverManager.getConnection(connURL);
if (id != null)
{
for( int i=0; i<id.length; i++){
String sqlStr = "Insert into assign(ArmID,Armnumber) Value(?,?)";
PreparedStatement pstmt = conn.prepareStatement(sqlStr);
pstmt.setInt(1, Integer.parseInt(id[i]));
pstmt.setString(2,armnumber);
//pstmt.setInt(1, Integer.parseInt(ids[i]));
int rec=pstmt.executeUpdate();
if (rec==1){
count++;
}
}
}
%>
<%
String[] id1 = request.getParameterValues("assignchkbox1");
String legID = request.getParameter("LegID");
String legnumber = request.getParameter("Legnumber");
int count1=0;
Connection conn1 = null;
try {
Class.forName("com.mysql.jdbc.Driver");
// Step 2: Define Connection URL
String conn1URL = "jdbc:mysql://localhost/medicloud?user=root&password=root";
// Step 3: Establish connection to URL
conn1 = DriverManager.getConnection(conn1URL);
if (id1 != null)
{
for( int i=0; i<id1.length; i++){
String sqlStr = "Insert into assign(LegID,Legnumber) Value(?,?)";
PreparedStatement pstmt = conn.prepareStatement(sqlStr);
pstmt.setInt(1, Integer.parseInt(id1[i]));
pstmt.setString(2,legnumber);
//pstmt.setInt(1, Integer.parseInt(ids[i]));
int rec=pstmt.executeUpdate();
if (rec==1)
count1++;
}
}
%>
<form action="assign.jsp" method="post">
<label><%=count%> arm record(s) assigned!!</label>
<td style width="50%">
</form>
<form action="assign.jsp" method="post">
<label><%=count1%> leg record(s) assigned!!</label>
<p><input type="submit" value="Return" name="ReturnBtn" /></p>
</form>
<%
conn.close();
conn1.close();
}finally{
}
}catch (Exception e) {
e.printStackTrace();
}
%>
</body>
</html>
暂无答案！
目前还没有任何答案，快来回答吧！